[codeforces1208D]Restore Permutation

time limit per test : 2 seconds
memory limit per test : 256 megabytes
An array of integers $p_1,p_2,…,p_n$ is called a permutation if it contains each number from $1$ to $n$ exactly once. For example, the following arrays are permutations: $[3,1,2],[1],[1,2,3,4,5]$and $[4,3,1,2]$. The following arrays are not permutations: $[2],[1,1],[2,3,4]$.

There is a hidden permutation of length $n$.

For each index $i$, you are given $s_i$, which equals to the sum of all $p_j$ such that $j<i$ and $p_j<p_i$. In other words, si is the sum of elements before the i-th element that are smaller than the $i$-th element.
Your task is to restore the permutation.

Input

The first line contains a single integer $n(1≤n≤2⋅10^5)$ — the size of the permutation.
The second line contains $n$ integers $s_1,s_2,…,s_n (0≤s_i≤n(n−1)/2)$.
It is guaranteed that the array $s$ corresponds to a valid permutation of length $n$.

Output

Print $n$ integers $p_1,p_2,…,p_n$— the elements of the restored permutation. We can show that the answer is always unique.

Examples
Input

3
0 0 0

Output

3 2 1

Input

2
0 1

Output

1 2

Input

5
0 1 1 1 10

Output

1 4 3 2 5

Note

In the first example for each $i$ there is no index j satisfying both conditions, hence $s_i$ are always 0.
In the second example for $i=2$ it happens that $j=1$ satisfies the conditions, so $s_2=p_1$.

In the third example for $i=2,3,4$ only $j=1$ satisfies the conditions, so $s_2=s_3=s_4=1$. For $i=5$ all $j=1,2,3,4$ are possible, so $s_5=p_1+p_2+p_3+p_4=10$.

题意：
给定一个序列 $\{s_n\}$， $s_i$= $\sum_{p_j<p_i且j<i} p_j$，要求生成一个 $n$的全排列 $\{ p_n \}$使其满足 $\{s_n\}$

题解：
从后往前构造解。
对于第 $i$个位置二分满足条件的 $p_i$，查询小于 $p_i$的和可以用树状数组来搞定。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline int lowbit(int x){return x&(-x);}
int n,a[200004];
ll s[200004];
ll tr[200004];
void add(int x,ll v){
    for(int i=x;i<=n;i+=lowbit(i))tr[i]+=v;
}
ll query(int x){
    ll ret=0;
    for(int i=x;i>=1;i-=lowbit(i))ret+=tr[i];
    return ret;
}
int Rget(ll x){
    int ret=-1;
    int l=1,r=n,mid;
    while(l<=r){
        mid=(l+r)>>1;
        ll pec=query(mid-1);
        if(query(mid)-pec!=0&&pec==x){
            ret=mid;
            add(ret,-ret);
            return ret;
        }
        if(pec<=x)l=mid+1;
        else r=mid-1;
    }
}
int main(){
    memset(tr,0,sizeof(tr));
    scanf("%d",&n);
    for(int i=1;i<=n;i++)add(i,i);
    for(int i=1;i<=n;i++)scanf("%lld",&s[i]);
    for(int i=n;i>=1;i--){
        a[i]=Rget(s[i]);
    }
    for(int i=1;i<=n;i++)printf("%d ",a[i]);
    puts("");
    return 0;
}